Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 14

An excited atom, of total mass \(m\), is at rest in a given frame. It emits a photon and thereby loses internal (i.e. rest) energy \(\Delta E\). Calculate the exact frequency of the photon, making due allowance for the recoil of the atom. [Answer: \((\Delta E / h)\left(1-\frac{1}{2} \Delta E / m c^{2}\right)\). Hint: use a four-vector argument.]

Short Answer

Expert verified
Answer: The exact frequency of the photon, considering the recoil, is given by the equation \(f = \frac{\Delta E}{h}\left(1- \frac{1}{2} \frac{\Delta E}{mc^2}\right)\), where ΔE is the internal energy lost by the atom, h is the Planck constant, m is the initial mass of the atom, and c is the speed of light.

Step by step solution

01

Introduction to four-vectors

In the context of special relativity, a four-vector is an object with four components that transform in a specific way under Lorentz transformations. The most common four-vector is the four-momentum, which is defined as \(P^{\mu} = (E/c, \vec{p})\), where E is the energy, c is the speed of light, and \(\vec{p}\) is the momentum vector. The dot product of two four-momenta is given by \(P^{\mu}Q_{\mu} = - P^0Q^0 + P^1Q^1 + P^2Q^2 + P^3Q^3\).
02

Conservation of four-momentum

The four-momentum is conserved in all processes in special relativity. In this case, we consider the emission of the photon by the atom. Before the emission, the four-momentum of the system (atom + photon) is \(P^{\mu}_{before} = (mc, 0, 0, 0)\), and after the emission, it can be written as \(P^{\mu}_{after} = (E_{atom}/c, \vec{p}_{atom}) + (E_{photon}/c, \vec{p}_{photon})\). The conservation of four-momentum requires \(P^{\mu}_{before} = P^{\mu}_{after}\).
03

Internal energy loss and mass change

The atom loses the internal energy \(\Delta E\), resulting in a decrease in its mass. The initial mass of the atom is \(m\) and the final mass is \(m' = m - (\Delta E/c^2)\), by using the mass-energy equivalence relation \(E=mc^2\). Thus, the four-momentum of the atom after the emission can be written as \(P^{\mu}_{atom} = (m'c, \vec{p}_{atom})\).
04

Calculate photon's four-momentum

Since the four-momentum is conserved, we can find the four-momentum of the photon by subtracting the atom's four-momentum before and after the emission: \(P^{\mu}_{photon} = P^{\mu}_{before} - P^{\mu}_{after} = ((\Delta E/c), \vec{p}_{photon})\). The photon's four-momentum is then given by \(P^{\mu}_{photon} = \left(\frac{\Delta E}{c}, \vec{p}_{photon}\right)\).
05

Determine the magnitude of photon's momentum

The square of the four-momentum is an invariant quantity, meaning that it remains unchanged under Lorentz transformations. Therefore, the square of the photon's four-momentum can be written as \(P^{\mu}_{photon}P_{\mu,photon} = 0\), since photons are massless. From this equation, we can determine the magnitude of the photon's momentum as \(p_{photon} = |\vec{p}_{photon}| = \frac{\Delta E}{c}\).
06

Calculate the frequency of the photon

The energy of a photon is related to its frequency by the equation \(E_{photon} = hf\), where \(h\) is the Planck constant, and \(f\) is the frequency of the photon. To find the exact frequency of the photon, considering the recoil, we can rearrange the equation for the photon's energy as \(f = \frac{\Delta E}{h}\). Finally, we can substitute the expression for the magnitude of the photon's momentum obtained from the four-momentum conservation to get the final answer: \(f = \frac{\Delta E}{h}\left(1- \frac{1}{2} \frac{\Delta E}{mc^2}\right)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If \(\Delta x^{*}=(c \Delta t, \Delta \mathbf{r})\) is the four-vector join of two events on the worldline of a uniformly moving particle (or photon), prove that the frequency vector of its de Broglie wave is given by $$ N^{\mu}=\frac{v}{c \Delta t} \Delta x^{\mu}, $$ whence \(v / \Delta t\) is invariant. Compare with Exercise III \((7)\).

taking \(h=6.63 \times 10^{-27} \mathrm{crgs}\) and \(c=300000 \mathrm{~km} \mathrm{~s}^{-1}\), calculate how many photons of wavelength \(5 \times 10^{-5} \mathrm{~cm}\) (in the yellowgreen of the visible spectrum) must fall per second on a blackened plate to produce a force of one dyne. [Answer: \(7.3 \times 10^{21}\). Hint: force equals momentum absorbed per unit time.]

Planck's constant \(h\) has the dimensions of action (energy \(x\) time or momentum \(\times\) distance) which suggests that the action of any periodic phenomenon may have to be a multiple of h. Accordingly Bohr constructed a model of the hydrogen atom in which the action of the single orbiting electron was quantized, requiring \(2 \pi r m v=n h\), \(n=1,2, \ldots\), where \(m\) is the mass of the electron, \(v\) its speed, and \(r\) the radius of the orbit. This led to a hydrogen spectrum which fitted the then known facts. Show that Bohr's hypothesis (1913) is equivalent to the assumption that a permissible orbit must contain an integral number of de Broglie electron waves.

15\. A rocket propels itself rectilinearly by emitting radiation in the direction opposite to its motion. If \(V\) is its final velocity relative to its initial rest frame, prove \(a b\) initio that the ratio of the initial to the final rest mass of the rocket is given by $$ \frac{M_{\mathrm{i}}}{M_{\mathrm{f}}}=\left(\frac{c+V}{c-V}\right)^{1 / 2} $$ and compare this with the result of Exercise 5 above. [Hint: equate energies and momenta at the beginning and at the end of the acceleration, writing \(\Sigma h v\) and \(\Sigma h v / c\) for the total energy and momentum, respectively, of the emitted photons.]

A particle with four-momentum \(P\) is observed by an observer who moves with four-velocity U \(_{0}\). Prove that the energy of the particle relative to that observer is \(\mathbf{U}_{0} \cdot \mathbf{P}\).
See all solutions

Recommended explanations on Physics Textbooks

Quantum Physics

Read Explanation

Space Physics

Read Explanation

Linear Momentum

Read Explanation

Dynamics

Read Explanation

Particle Model of Matter

Read Explanation

Energy Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free